ZJSG09112007
Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

I need help with these questions (see image). Please show workings.​

I Need Help With These Questions See Image Please Show Workings class=

Sagot :

Answer:

r = [tex]\frac{1}{2}[/tex] [tex]\sqrt{l^2+4h^2}[/tex]

Step-by-step explanation:

h bisects l at right angles.

There is a right triangle formed with legs h and [tex]\frac{1}{2}[/tex] l and hypotenuse r

Using Pythagoras' identity, then

r² = ([tex]\frac{1}{2}[/tex] l )² + h²

   = [tex]\frac{1}{4}[/tex] l² + h² ← factor out [tex]\frac{1}{4}[/tex]

    = [tex]\frac{1}{4}[/tex] (l² + 4h²)

Take the square root of both sides

r = [tex]\sqrt{\frac{1}{4}(l^2+4h^2) }[/tex] = [tex]\frac{1}{2}[/tex] [tex]\sqrt{l^2+4h^2}[/tex]

mhanifa

Answer:

  • r = [tex]\sqrt{(l/2)^2 + h^2}[/tex]

Step-by-step explanation:

Connect the endpoints of the with the center. The formed triangle is isosceles with sides l, r and r.

The height of the triangle h, is the distance between the center of the circle and the midpoint of the chord.

As per Pythagorean theorem we have:

  • (l/2)² + h² = r²
  • [tex]\sqrt{(l/2)^2 + h^2}[/tex] = r
  • r = [tex]\sqrt{(l/2)^2 + h^2}[/tex]

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.